Algebra and Geometry of Sets in Boolean Space

In the present paper, geometry of the Boolean space Bn in terms of Hausdorff distances between subsets and subset sums is investigated. The main results are the algebraic and analytical expressions for representing of classical figures in Bn and the functions of distances between them. In particular, equations in sets are considered and their interpretations in combinatory terms are given.


Distance between Subsets B n
( ) ( ) , and ⊕ is the addition operation with respect to mod 2. The Hausdorff distance has essential role in many problems of discrete analysis [1] and thus has certain interest.On the other hand, there only are a few essential results concerning distances between the subsets n B , and their investigation offers significant difficulties.First, we present the following simple properties of the Hausdorff distance: X a Y a ρ ρ = + + , for n a B ∈ .Let us note that, generally speaking, the Hausdorff distance does not satisfy the triangle inequality: which is demonstrated in the following picture: But inequality (1) holds true if 1 Z = .
We present them in the form: ( ) , min min min max .
x M y M x M y M x M y M x p y q x x y y x y x y x y x y Then, taking into account that:

max ,
x p y q x y p q p q n we have: , , .
The theorem is proved.Let: ( ) ( ) Taking into account that the sphere of the radius p with the center at ( ) 00 0  and the sphere of the radius q with the center at ( )  contain, respectively, as many points as: 0 and , we get the following corollary.
The value of the function ( ) , R r r for definite values of 1 2 , r r was calculated in [1].Theorem 2. If q p > , then: 0 , .
The general form of the standard generating function for the distance between the subsets , n X Y B ⊆ has the following form: The summation in ( 2) is over all pairs of the subsets ( ) In this case, we have: .
, which is the well-known function of distribution of distances between the points in the space n B with the metrics of Hamming.2) 1 p = , and q is an arbitrary positive integer which does not exeed 2 n .In this case: As: ( ) ( ) ( ) Consequently, the distance between the zero point and an arbitrary subset Y equals the minimal weight of the points which are in Y.
Hence, ( ) ( ) is the cardinality of the sphere with the radius r in .
n B From (5), we get the following statement: is the number of the subsets of cardinality q (in n B ) having the distance to the zero point, then: The following formula holds true: ( ) Proof.By definition: ( ) From this and Lemma 1, taking into account (3) and ( 4), we get: Φ p z is the generating function of the random value ( ) uniformly distributed on the pairs ( ) , a Y , where Y q = , then the following holds true.Corollary 1.The following formula holds true: The following holds true: F z follows from (6).Proof.From (6) we get for p = 1: Corollary 4. For q = 2, the following formula holds true: ( ) By definition and from Corollary 2, we derive: ( ) Transforming the terms in (7), we get: Then, using the following formulas: Let us "compress" the sum: By definition, we have: ( ) And: ( ) Taking this and (8) into account, we get: ( ) And the generating function: can be expressed by the following parameters: then the family of all subsets of cardinality p, having distances r ≥ from M is expressed as . Indeed, ( ) does not contain such points; consequently, for an arbitrary subset ( ) , the expression ( ) The cardinality of this family is: 2) The number of all m-element subsets having the distance r from M is: Summarizing all the previous, we get the following statement.Theorem 4. The following expression is true: .
The operation "+" is defined in the family 2 Here both limits are reachable.The properties of "+" are as follows: 1) 0 ; + Thus, there is certain analogy between the norm of the sum of points and the distance between those points, as well as between the norm of the sum of the sets and the distance between those sets.
In the general form, the following statement connecting the operations "∪" и+, is true:

Sum of Facets in B n and the Distance between Them
A facet or interval in n B is the set of points { }, where the partial order x y ≤ is defined in the classic way [5] [6]: , 1, , , .then every point of J can be presented by the word ( ) which means the following: all the points which are obtained from the word ( ) 0 1 cc by the substitution either 0 or 1 for a letter of the given word, are contained in the interval J. Consequently, the cardinality of the interval J is 2  2 4, = for the given case, i.e.

J =
Hence, each interval J has its corresponding code word ( ) J λ in the alphabet A. The number of letters c in the code ( ) is the dimension of the interval J, i.e. is dim .J And the following formula is obvious: dim 2 .

J J =
If the operation "*" is introduced on the alphabet A: Examples.
for every interval 2 .J The distance between the intervals 1 J and 2 J , having the codes ( ) , .J J J J ρ λ = + Thus, the distance between the intervals 1 J and 2 J is the number of occurrences of letter 1 in the code of their sum. Examples.
and introduce uniform distribution on it with the generating function: The following formula is true: ∮ where 1. r < Let us consider the matrix i j α the rows of which are the codes ( ) ( ) ( ) , , , .
The following expression is true: where 0 1 i i k k is the number of zeros and, respectively, units in the i-th column of the matrix .
, , where 1, if and , ; 0, otherwise, and ( ) ( ) is the code of the interval i J .It follows from ( 9): The internal sum in (10) equals the number of such pairs ( ) be the family of all edges in 3 B .We consider the matrix of their codes: The total number of the edges in 3 B is

∕
. Each column of the matrix has the length 12, and all letters of the alphabet { } 0,1, c occur in equal number, 4 times.Therefore, 0 From here, we get:  ( ) From here, we get: Taking this into account, to describe the set , B B + it is sufficient to describe only the points which are included into this sum.The minimal weight of these is d p q = − .We discuss the following outline: and the "block" of the first 2 units is shifted by a unit in each consecutive word.Thus, we get all weights: : 2, 4, 6.
i z z + In the general case, the situation is absolutely analogous, and the weights are arranged as follows: , 2, 4, , p q p q p q − − + − +  for .p q ≥ Here the condition for z holds true: , 2 p q r p q n p q − + ≤ + − + . Examples.
min , , max 0, 1. l p q n l p q = + = − − Proof.From Lemmas 3 and 5, we have: The proof is over.

Sum of Subspaces in B n
As usual, let ( ) L X be the subspace generated by the vectors from the set X, or be the space "worn" on X.
Proof.It follows from (15) that it is sufficient to prove that the following equality is true: , X Y be a solution of the equation , X Y A + = for which: On the other hand, it follows from Statement 11 that ( ) Taking this and (16) into account, we get: ( ) ( ).
distance between the points: a sphere of radius p with the center at n x B ∈ .We take, for an arbitrary subset,

≠
the set of all points in n B which are at the distance p ≤ from the set М; the set of all points in n B , covered by the spheres of the radius p with the centers at points of the set M. then p is the radius of the covering of the set n B[2].Theorem 1.
+") is a monoid with the neutral element 0 0 n = [3] [4].Besides, the following inequality holds true: Every interval J can be written in the form of a word of the length n, in the alphabet account the introduced definitions-are calculated in the following way.Let ( )1 J λ be the number of occurrences of letter 1 in the code of the interval J.
According to the definition:

2 ) 1 2
In the general form, if ( ) S n is the family of all edges in n B , each column contains n− letters c and .